$\vec w = (-2,2)$ $-2\vec w= ($
Solution: In general, the scalar multiple of $k$ times $\vec u$ is this: $k\vec u = k(u_x, u_y) = (ku_x, ku_y)$. So, here's how we find $-2 \vec{w}$ : $\begin{aligned} {-2}\vec w = {-2} \cdot (-2,2) &= \left({-2} \cdot (-2), {-2} \cdot 2\right) \\\\ &= (4,-4) \end{aligned}$ The answer is $ (4,-4) $.